The number of solutions of the equation , is (where denotes greatest integer function and fractional part function)

JEE MainMathematicsToughITF

The number of solutions of the
equation $\ \tan^{- 1}(4\{ x\}) + \cot^{- 1}(x + \lbrack x\rbrack)) = \frac{\pi}{2}$ ,
is (where denotes greatest integer function and fractional
part function)

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\because 4\{ x\} = x + \lbrack x\rbrack\ldots(1)

Case-1 Let x=I

O = I + I \Rightarrow I = 0

Case-2 Let

x = I + F,0 < f < 1

Equation (1) becomes

4f = I + f + I

f = \frac{2I}{3}\therefore I = 1,f = \frac{2}{3}\therefore r = \frac{5}{3}

\therefore

two solutions

Source: Grass-G / M-17 Mathematics Paper 26 Nov Second.docx

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